Suppose that $f: \mathbb S^1\to \mathbb C$ is continuous such that $f(z)=f(\bar z)$ for all $z\in \mathbb S^1$. Can it be extended to a continuous $F: \overline{\mathbb D}\to \mathbb C$ such that $ F$ is holomorphic on $\mathbb D$?
If $f$ is constant, then the result is obviously true.
So suppose that $f$ is not constant. Suppose that such an extension $F$ exists. Take any $r\in (0.2,1)$. For any $z\in D_{1/10}(0)$, we have $$F(z)=\int_{|z|=r}F(u)/(u-z) \,du\tag 1$$
$(1)$ gives $F(z)=\sum_{k\ge 0}\left(\int_{|z|=r}\frac{F(u)}{u^{k+1}}\right)z^k=\sum_{k\ge 0} a_k(r) z^k$, where $(r)$ denotes that $a_k$ depends upon $r$. Calculations show that $2\pi r^ka_k(r)=\color{blue}{\int_0^{2\pi} F(re^{-i\theta})e^{ki\theta} \,d\theta}$.
By continuity of $F,F(re^{-i\theta})\overbrace{\to}^{r\to 1-} F(e^{-i\theta})$ $$\implies 2\pi a_k(r)\overbrace{\to}^{r\to 1-}\int_0^{2\pi} \color{blue}{F(e^{-i\theta})}e^{ki\theta} \,d\theta=\int_0^{2\pi} \color{blue}{F(e^{i\theta})}e^{ki\theta} \,d\theta=i\int_{|z|=1}F(z)z^{k-1}dz=0\,\forall k\ge 1$$
It follows that $a_k(r)\to 0$ as $r\to 1-$.
$\color{red}{\text{Since $z$ is fixed and power series representation is unique, we must have $a_k(r)=$ constant.}}$ Hence $a_k(r)=0$ for all $k\ge 1$.
It follows that $F$ is a constant in $D_{1/10}(0)$, whence by identity theorem it follows that $F$ is constant which contradicts our assumption. So such an extension is not possible.
Is this correct? I am not sure about validity of the read part. Can anyone please help me justify or disprove the red colored part? Thanks.
I think the idea is correct. I would argue the following way: Let $F:\overline{\mathbb{D}} \to \mathbb{C}$ be a continuous extension of $f$ which is holomorphic on $\mathbb{D}$. Then $F$ can be represented by a power series on $\mathbb{D}$: $$ F(z)=\sum_{n=0}^\infty a_nz^n \quad (z \in \mathbb{D}). $$ For each $k \in \mathbb{Z}$ and $r \in (0,1)$ $$ \frac{1}{2\pi}\int_0^{2\pi} F(re^{it})e^{-ikt}dt = \sum_{n=0}^\infty \left(a_nr^n \frac{1}{2\pi}\int_0^{2\pi} e^{i(n-k)t} dt \right)= \left\{\begin{array}{cc} a_kr^k, & k \in \mathbb{N}_0 \\ 0, & k \in -\mathbb{N} \end{array}\right. $$ As you noted $F(re^{it}) \to F(e^{it})$ $(r \to 1-)$ (uniformly on $[0,2\pi]$). Thus $$ \frac{1}{2\pi}\int_0^{2\pi} F(e^{it})e^{-ikt}dt =\left\{\begin{array}{cc} a_k, & k \in \mathbb{N}_0 \\ 0, & k \in -\mathbb{N} \end{array}\right. $$ Now, let $k \in \mathbb{N}$ and substitute $s=-t$: $$ 2 \pi a_k=\int_0^{2\pi} F(e^{it})e^{-ikt}dt = \int_0^{2\pi} F(e^{-it})e^{-ikt}dt = -\int_0^{-2\pi} F(e^{is})e^{iks}ds $$ $$ =\int_{-2\pi}^0 F(e^{is})e^{iks}ds = \int_{0}^{2\pi} F(e^{is})e^{iks}ds = 0, $$ since $-k \in -\mathbb{N}$. Thus $a_k =0$ $(k \in \mathbb{N})$.